Find the zeros for f(x)= (x^2-16x+64)(x-2)

Question

anonymous · Answer

I can help

anonymous · Answer

f(x) = x² + 16x + 62 

Zeros: f(x) = 0: 

x² + 16x + 62 = 0 
x² + 16x = - 62 
x² + 16x + 64 = 64 - 62 
(x + 8)² = 2 
x + 8 = √2 
x = - 8 ± √2 
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anonymous · Answer

Source:
3/11/14

mathstudent55 · Answer

@saynabahmed13 
The problem is:
(x^2-16x+64)(x-2) = 0
You used the wrong trinomial.

campbell_st · Answer

you have been given a linear factor and a quadratic factor 

$$f(x) = (x^2 - 16x + 64)(x - 2)$$

the quadratic factor is a perfect square

$$x^2 - 2xb + b^2 = (x - b)^2$$

can you calculate b?

campbell_st · Answer

the linear factor 

(x - 2) to find the zero, let it equal zero and solve for x

so

x - 2 = 0

what value does x take...?

campbell_st · Answer

with factoring the perfect square, you need to compare

$$x^2 - 16x + 64 ~~~and x^2 -2xb + b^2 $$

to find b so it can be written in the factored form 

$$(x - b)^2$$